Metamath Proof Explorer
Description: The absolute value of a difference and 'less than' relation.
(Contributed by Mario Carneiro, 29-May-2016)
|
|
Ref |
Expression |
|
Hypotheses |
absltd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
|
|
absltd.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ ) |
|
|
absltd.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℝ ) |
|
Assertion |
absdifltd |
⊢ ( 𝜑 → ( ( abs ‘ ( 𝐴 − 𝐵 ) ) < 𝐶 ↔ ( ( 𝐵 − 𝐶 ) < 𝐴 ∧ 𝐴 < ( 𝐵 + 𝐶 ) ) ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
absltd.1 |
⊢ ( 𝜑 → 𝐴 ∈ ℝ ) |
2 |
|
absltd.2 |
⊢ ( 𝜑 → 𝐵 ∈ ℝ ) |
3 |
|
absltd.3 |
⊢ ( 𝜑 → 𝐶 ∈ ℝ ) |
4 |
|
absdiflt |
⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( abs ‘ ( 𝐴 − 𝐵 ) ) < 𝐶 ↔ ( ( 𝐵 − 𝐶 ) < 𝐴 ∧ 𝐴 < ( 𝐵 + 𝐶 ) ) ) ) |
5 |
1 2 3 4
|
syl3anc |
⊢ ( 𝜑 → ( ( abs ‘ ( 𝐴 − 𝐵 ) ) < 𝐶 ↔ ( ( 𝐵 − 𝐶 ) < 𝐴 ∧ 𝐴 < ( 𝐵 + 𝐶 ) ) ) ) |